I have the definition of a homomorphism as map such that $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$
I have the definition of $\mathrm{Hom}(V,W)$ as
$$\begin{align}\mathrm{Hom}(V,W) &= \{\mathbb{C}-\text{linear maps }\varphi:V \rightarrow W \} \\ &\cong \{n \times m \text{ matrices } \} \end{align}$$
Does $\mathrm{Hom}$ stand for homomorphisms because I cannot see how it relates to the definition of homomorphism
I have the definition of $\mathrm{Hom}_G$ as
$$\begin{align}\mathrm{Hom}_G(V,W) &= \{\varphi \in \mathrm{Hom}(V,W) \mid g\varphi(v)=\varphi(gv), \forall g \in G, \forall v \in V\} \\ &=\{\varphi \in \mathrm{Hom}(V,W) \mid \rho(g)\cdot\varphi(v)=\varphi(\rho(g)v) , \forall g \in G, \forall v \in V\} \end{align}$$
I then have the question:
Prove that if $\rho$ is an irreducible representation, then for an element $g \in G$
$$g \in Z(G) \iff \rho(g)=\lambda I$$
In the solution I have that:
$(``\implies")$ If $g \in Z(G)$ then $gh=hg \ \forall h \in G$. By definition of $\mathrm{Hom}_G$ this means that $\rho(g) \in \mathrm{Hom}_G(V,V)$. But \rho is irreducible so $\mathrm{Hom}_G(V,V)$ consists of scalar matrices by Schur's Lemma.
I do not understand how "By definition of $\mathrm{Hom}_G$ this means that $\rho(g) \in \mathrm{Hom}_G(V,V)$. " How is this the definiton of $\mathrm{Hom}_G$? I cannot see why this is equivalent.
The word homomorphism is generally used to mean "a map that preserves the structure of the object I'm looking at". Thus the precise definition is different for groups, vector spaces, group representations, etc., even though we use the same word.
For your second question, assume that $g$ is an element of the center of $G$, and let $\varphi = \rho(g)$. Let $h$ be an element of $G$. Since $g$ is in the center, we have $gh=hg$, so $\rho(g)\cdot \rho(h) = \rho(h)\cdot\rho(g)$. By our notation, this can be written as $\rho(h)\cdot\varphi = \varphi\cdot\rho(h)$; in other words, for any $v\in V$, we have that $\rho(h)\cdot\varphi(v) = \varphi(\rho(h)(v))$. By definition of $\mathrm{Hom}_G$, this means that $\varphi\in \mathrm{Hom}_G(V,V)$. Since $\varphi=\rho(g)$, this answers your question.